/* determine minimal prime 3x3 magic square; for more details see bottom */

#include <stdio.h>
#include <sys/time.h>
#include <stdint.h>

uint32_t B[]={0x35145105,0x4510414,0x11411040,0x45144001};

#define Prime(i) ((B[(i)>>5] & (0x80000000UL >> ((i)%32))) != 0)

#define forall_odd_primes_less_than(p, m, block) \
  for((p)=3; (p)<(m); (p)+=2)                    \
    if (Prime((p)))                              \
      block

uint8_t p,a,b,c,d;
struct timeval tv0,tv1;

int main(void)
{
  gettimeofday(&tv1, NULL);      // wait for usec change
  do  gettimeofday(&tv0, NULL);  while (tv0.tv_usec == tv1.tv_usec);

  forall_odd_primes_less_than(p, 64,
    forall_odd_primes_less_than(a, p,
      if Prime(2*p-a)
      {
        forall_odd_primes_less_than(b, p,
          if ( (b!=a) && Prime(2*p-b) )
          {
            c= 3*p - (a+b);

            if ( (c<2*p) && (2*p-c!=a) && (2*p-c!=b) && Prime(c) && Prime(2*p-c) )
            {
              if (2*a+b>2*p)
              {
                d = 2*a + b - 2*p;   // 3*p - (3*p-(a+b)) - (2*p-a)

                if ( (d!=a) && (d!=b) && (d!=2*p-c) && Prime(d) && Prime(2*p-d) )
                {
                  gettimeofday(&tv1, NULL);

                  printf("%3u|%3u|%3u|\n%3u|%3u|%3u|\n%3u|%3u|%3u|\n",
                    a,b,c,2*p-d,p,d,2*p-c,2*p-b,2*p-a);

                  printf("\n%ldus\n",
                    1000000*(tv1.tv_sec-tv0.tv_sec)+tv1.tv_usec-tv0.tv_usec);
                  return 0;
                }
              }
            }
          }
        )
      }
    )
  )
}

/*

it always exists this by rotation and flippings (= is p, -/+ is less/greater p)
--?
?=?
???

proof by enumeration of all possibilities

++
 =
 --

+-  +-+ +-+
 =   =  +=-
 +- -+- -+-

        +-+
        -=
        -+-

    +--
     =
     +-

-+  -+  -++
 =  +=- +=-
 -+  -+ --+

        -+-
        +=-
         -+

    -+
    -=
     -+

--
 =
 ++



row/column/diagonal sum is 3*p

a b 3*p-(a+b)=c   - - +

  p 2*a+b-2*p=d   + = -

    2*p-a         - + +
*/